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4 Laplace equation

4.1 Ill-posedness of Cauchy problem for Laplace equation

In the study of various classes of solutions to the Cauchy problem for the wave equation wewere able to establish

• existence of the solution in a suitable class of functions;

• uniqueness of the solution;

• continuous dependence of the solution on the initial data with respect to a suitabletopology.One may ask whether these properties remain valid for other evolutionary PDEs?

Changing the sign in the wave equation one arrives at the Laplace equation of elliptictype, which in R2 reads

utt + a2uxx = 0. (4.1.1)

Does the change of the type of equation affect seriously the properties of solutions?

To be more specific we will deal with the periodic Cauchy problem

u(x, 0) = φ(x), ut(x, 0) = ψ(x) (4.1.2)

with two 2π-periodic smooth initial functions φ(x), ψ(x). For simplicity let us choose a = 1.We will see that the solution to this Cauchy problem does not depend continuously on theinitial data. For that let us consider the sequence of solutions uk(x, t), k ∈ N to the Cauchyproblem with following initial data:

u(x, 0) = 0, ut(x, 0) =sin k x

k. (4.1.3)

The 2π-periodic solution (if any) can be expanded in Fourier series

uk(x, t) =a0(t)

2+∞∑n=1

[an(t) cosnx+ bn(t) sinnx]

with some coefficients an(t), bn(t). Substituting the series into equation

utt + uxx = 0

we obtain an infinite system of ODEs

an = n2an, ∀n ∈ Nbn = n2bn, ∀n ∈ N \ 0.

The initial data for this infinite system of ODEs follow from the Cauchy problem (4.1.2):

an(0) = 0, an = 0,

bn(0) = 0, bn(0) =

1/k, n = k0, n 6= k.

The solution has the form

an(t) = 0 ∀n, bn(t) = 0 ∀n 6= k

bk(t) =1

k2sinh kt.

So the solution to the Cauchy problem (4.1.2) reads

uk(x, t) =1

k2sin kx sinh kt. (4.1.4)

Using this explicit solution we can prove the following

46

Preliminary version – December 2, 2012

Theorem 4.1.1 (J.Hadamard (1922)). For any positive ε, M , t0 there exists an integer Ksuch that for any k ≥ K the initial data (4.1.3) satisfy

supx∈[0,2π]

(|uk(x, 0)|+ |∂tuk(x, 0)|) < ε (4.1.5)

but the solution uk(x, t) at the moment t = t0 > 0 satisfies

supx∈[0,2π]

(|uk(x, t0)|+ |∂tuk(x, t0)|) ≥M. (4.1.6)

Proof: Since l.h.s. of (4.1.5) is smaller that 1k , choosing an integer K1 satisfying

K1 >1

ε

we will have the inequality (4.1.5) for any k ≥ K1. In order to obtain a lower estimate of theform (4.1.6) let us first observe that

supx∈[0,2π]

(|uk(x, t0)|+ |∂tuk(x, t0)|) =1

k2sinh kt0 +

1

kcosh kt0 >

ekt0

k2

where we have used an obvious inequality

1

k>

1

k2for k > 1.

The last function of k is monotone increasing for k > 2t0

and grows to∞, hence for any t0 > 0and M > 0 there exists K2 such that for any k > K2

ek t0

k2> M.

ChoosingK = max(K1,K2)

we complete the proof.

The statement of the Theorem is usually referred to as ill-posedness of the Cauchy problem(4.1.1), (4.1.2).

A natural question arises: what kind of initial or boundary conditions can be chosenin order to uniquely specify solutions to Laplace equation without violating the continuousdependence of the solutions on the boundary/initial conditions?

4.2 Dirichlet and Neumann problems for Laplace equation on the plane

We consider the (elliptic) Laplace operator in the d-dimensional Euclidean space

∆ =∂2

∂x21

+ · · ·+ ∂2

∂x2d

(4.2.1)

and the Laplace equation in an open domain Ω

∆u(x) = 0, x = (x1, . . . xd) ∈ Ω ⊂ Rd. (4.2.2)

The (real) solutions to the Laplace equation are called harmonic functions in Ω.

47


We will formulate the two main boundary value problems (b.v.p.’s), assuming that theboundary ∂Ω of the domain Ω is a smooth hypersurface. Moreover we assume that thedomain Ω does not go to infinity, i.e. Ω belongs to some ball in Rd.

Problem 1 (Dirichlet problem). Given a function f(x) defined at the points of theboundary find a function u = u(x) satisfying the Laplace equation on the internal part of thedomain Ω and the boundary condition

u(x)|x∈∂Ω = f(x) (4.2.3)

on the boundary of the domain.

Problem 2 (Neumann problem). Given a function g(x) defined at the points of theboundary find a function u = u(x) satisfying the Laplace equation on the internal part of thedomain Ω and the boundary condition(

∂u(x)

∂n

)x∈∂Ω

= g(x) (4.2.4)

on the boundary of the domain. Here n = n(x) is the unit external normal vector at everypoint x ∈ ∂Ω of the boundary and the normal derivative is defined as

∂u(x)

∂n:= n · ∇u = n1

∂u(x)

∂x1+ · · ·+ nd

∂u(x)

∂xd. (4.2.5)

Recall that the components ni of n are just equal to the cosαi, where αi are the anglesbetween n and the coordinate axes ei.

Example 1. For d = 1 the Laplace operator is just the second derivative

∆ =d2

dx2.

The Dirichlet b.v.p. in the domain Ω = (a, b), ∂Ω = a, b,

u′′(x) = 0, u(a) = fa, u(b) = fb

has an obvious unique solution

u(x) =fb − fab− a

(x− a) + fa.

The Neumann b.v.p. in the same domain, since n = n1∂1 with n1(a) = −1 and n1(b) = 1, is

u′′(x) = 0, −u′(a) = ga, u′(b) = gb,

and so its solution isu(x) = gbx+ c,

where c is an arbitrary constant, but only provided that

ga + gb = 0 (4.2.6)

since the derivatives of such u(x) are equal at a and at b

u′(a) = gb = u′(b).

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Lemma 4.2.1. In two dimensions the Laplace operator

∆ =∂2

∂x2+

∂2

∂y2. (4.2.7)

in the polar coordinatesx = r cosφy = r sinφ

(4.2.8)

takes the form

∆ =∂2

∂r2+

1

r

∂

∂r+

1

r2

∂2

∂φ2. (4.2.9)

Proof: We have (for brevity we denote c = cosφ, s = sinφ)

∂r = xr∂x + yr∂y = c∂x + s∂y (4.2.10)

∂φ = −rs∂x + rc∂y.

Hence∂2r = c2∂2

x + s2∂2y + 2sc∂x∂y

and∂2φ = ∂φ(−rs∂x + rc∂y) = ∂φ(−rs)∂x + ∂φ(rc)∂y − rs∂φ∂x + rc∂φ∂y

= −rc∂x − rs∂y + r2s2∂2x + r2c2∂2

y − 2r2sc∂x∂y.

Then a simple computation gives the result.

In the particular case of a disc of radius ρ

Ω = B(0, ρ) = (x, y) |x2 + y2 < ρ2 = (r, φ) | 0 ≤ r < ρ (4.2.11)

the Dirichlet b.v.p. is formulated as follows: find a solution u = u(x, y) to the Laplaceequation in Ω satisfying the boundary condition

u(r, φ)|r=ρ = f(φ). (4.2.12)

Here we represent the boundary condition defined on the boundary of the disc as a functiondepending only on the polar angle φ. Similarly, the Neumann problem consists of finding asolution to the Laplace equation satisfying

∂u(r, φ)

∂r|r=ρ = g(φ) (4.2.13)

for a given function g(φ). Indeed, the partial derivative ∂r (4.2.10), is the normal derivative(note that sinφ = cos(π2−φ) and φ and π

2−φ are angles with the axes x1 and x2, respectively).

Let us return to the general d-dimensional case. The following identity will be useful inthe study of harmonic functions.

Theorem (Green’s formula). For arbitrary smooth functions u, v on a bounded closeddomain Ω with a piecewise smooth boundary ∂Ω∫

Ω∇u · ∇v dV +

∫Ωu∆v dV =

∫∂Ωu∂v

∂ndS. (4.2.14)

Here

∇u · ∇v =

d∑i=1

∂u

∂xi

∂v

∂xi

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is the inner product of the gradients of the functions,

dV = dx1 . . . dxd

is the Euclidean volume element, n the external normal and dS is the area element on thehypersurface ∂Ω.

For proof see e.g. [Evans, Appendix C].

Example 1. For d = 1 and Ω = (a, b) the Green’s formula reads∫ b

aux vx dx+

∫ b

au vxx dx = u vx|ba

since the oriented boundary of the interval consists of two points ∂[a, b] = b − a. This is aneasy consequence of integration by parts.

Example 2. For d = 2 and a rectangle Ω = (a, b)× (c, d) we show the Green’s formula∫Ω

(uxvx + uyvy) dx dy +

∫Ωu (vxx + vyy) dx dy

=

∫ d

c

∫ b

a(u vx)x dx dy +

∫ b

a

∫ d

c(u vy)y dy dx

=

∫ d

c(u vx) |x=b

x=a dy +

∫ b

a(u vy) |y=d

y=c dx.

As it should, the right hand side of this formula is a sum (with a sign) of integrals over thefour pieces of the boundary ∂Ω.

Remark. These two examples essentially ’prove’ (4.2.14) since we can approximate any Ωby a union of rectangulars, and we can analogously proceed in higher dimensions.

Corollary 4.2.2. The Green’s formula∫Ω

∆v dV =

∫∂Ω

∂v

∂ndS. (4.2.15)

we used in Sect 3.8 follows by substitution u = 1 in (4.2.14).

Corollary 4.2.3. For a harmonic function u in a domain Ω with a piecewise smooth boundary∂Ω the integral of the normal derivative of u over ∂Ω vanishes:∫

∂Ω

∂u

∂ndS = 0 (4.2.16)

(use (4.2.15)). Moreover,∫Ω

(∇u)2 =

∫∂Ωu∂nu dS =

∫∂Ω

1

2∂nu

2 dS (4.2.17)

(substitute u = v in (4.2.14)).

Using the last identity we can easily derive the following uniqueness properties of solutionto the Dirichlet end Neumann boundary problems.

Theorem 4.2.4. 1) Let u1, u2 be two functions harmonic in the domain Ω, smooth in theclosed domain Ω, and coinciding on the boundary ∂Ω. Then u1 ≡ u2.

2) Under the same assumptions about the functions u1, u2, if the normal derivatives onthe boundary coincide

∂u1

∂n=∂u2

∂nthen the functions differ by a constant.

50


Proof: Applying to the difference u = u2 − u1 the identity (4.2.17) one obtains∫Ω

(∇u)2dV = 0

since the right hand side vanishes. Hence ∇u = 0, and thus the function u is equal to aconstant. The value of this constant on the boundary is zero. Therefore u ≡ 0. The secondstatement has a similar proof.

The following counterexample shows that the uniqueness does not hold true for infinitedomains. Let Ω be the upper half plane:

Ω = (x, y) ∈ R2 | y > 0..

The linear function u(x, y) = y is harmonic in Ω and vanishes on the boundary. Clearly u isdifferent from 0, which is obviously harmonic too.

Our goal is to solve the Dirichlet and Neumann boundary value problems. The first resultin this direction is the following

Theorem 4.2.5. For an arbitrary C1-smooth 2π-periodic function f(φ) the solution to theDirichlet b.v.p. (4.2.12) in the disc Ω = B(0, ρ) exists and is unique. Moreover it is given bythe following formula

u(r, φ) =1

2π

∫ 2π

0

ρ2 − r2

ρ2 − 2ρ r cos(φ− ψ) + r2f(ψ) dψ. (4.2.18)

The expression (4.2.18) for the solution to the Dirichlet b.v.p. is called Poisson formula.

Proof: We will first use the method of separation of variables in order to construct particularsolutions to the Laplace equation. At the second step we will represent solutions to theDirichlet b.v.p. as a linear combination of the particular solutions.

The method of separation of variables starts from looking for solutions to the Laplaceequation in the form

u = R(r)Φ(φ). (4.2.19)

Here r, φ are the polar coordinates on the plane (see Exercise 4.2.1 above). Using the form(4.2.9) we reduce the Laplace equation ∆u = 0 to

R′′(r)Φ(φ) +1

rR′(r)Φ(φ) +

1

r2R(r)Φ′′(φ) = 0.

After division by 1r2R(r)Φ(φ) we can rewrite the last equation in the form

R′′(r) + 1rR′(r)

1r2R(r)

= −Φ′′(φ)

Φ(φ).

The left hand side of this equation is independent of φ while the right hand side is independentof r. The equality is possible only if both sides are equal to some constant λ. In this way wearrive at two ODEs for the functions R = R(r) and Φ = Φ(φ)

R′′ +1

rR′ − λ

r2R = 0 (4.2.20)

Φ′′ + λΦ = 0. (4.2.21)

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We have now to determine the admissible values of the parameter λ. To this end let us beginfrom the second equation (4.2.21). Its solutions have the form

Φ(φ) =

Ae√−λφ +B e−

√−λφ, λ < 0

A+B φ, λ = 0

A cos√λφ+B sin

√λφ, λ > 0

.

Since the pairs of polar coordinates (r, φ) and (r, φ+2π) correspond to the same point on theEuclidean plane the solution Φ(φ) must be a 2π-periodic function. Hence we must discardthe negative values of λ and we have B = 0 if λ = 0. Moreover λ must have the form

λ = n2, n = 0, 1, 2, . . . . (4.2.22)

This givesΦ(φ) = A cosnφ+B sinnφ (4.2.23)

(λ = n = 0 corresponds to a constant Φ(φ) = A).The first ODE (4.2.20) for λ = n2 becomes

R′′ +1

rR′ − n2

r2R = 0.

This is a particular case of Euler equation. One can look for solutions in the form

R(r) = rk.

The exponent k has to be determined from the characteristic equation

k(k − 1) + k − n2 = 0

obtained by the direct substitution of R = rk into the equation. The roots of the characteristicequation are k = ±n. For n > 0 this gives the general solution of the equation (4.2.20) inthe form

R = a rn +b

rn

with two integration constants a and b. For n = 0 the general solution is

R = a+ b log r.

As the solution must be smooth at r = 0 one must always choose b = 0 for all n. In this waywe arrive at the following family of particular solutions to the Laplace equation

un(r, φ) = rn (An cosnφ+Bn sinnφ) , n = 0, 1, 2, . . . (4.2.24)

We want now to represent arbitrary solution to the Dirichlet b.v.p. in the disc of radius ρ asa linear combination of these solutions:

u(r, φ) =A0

2+∑n≥1

rn (An cosnφ+Bn sinnφ)

(4.2.25)

u(r, φ)|r=ρ = f(φ).

The boundary data function f(φ) must be a 2π-periodic function. Assuming this function tobe C1-smooth let us expand it in Fourier series

f(φ) =a0

2+∑n≥1

(an cosnφ+ bn sinnφ)

an =1

π

∫ 2π

0f(φ) cosnφdφ, bn =

1

π

∫ 2π

0f(φ) sinnφdφ. (4.2.26)

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Comparison of (4.2.25) with (4.2.26) yields

An =anρn, Bn =

bnρn,

or, equivalently

u(r, φ) =a0

2+∑n≥1

(r

ρ

)n(an cosnφ+ bn sinnφ) . (4.2.27)

Recall that this formula holds true on the disc of radius ρ, i.e., for r ≤ ρ.Taking into accout the expression for an, bn the last formula can be rewritten as follows:

u(r, φ) =1

π

∫ 2π

0

1

2+∑n≥1

(r

ρ

)n(cosnφ cosnψ + sinnφ sinnψ)

f(ψ) dψ

=1

π

∫ 2π

0

1

2+∑n≥1

(r

ρ

)ncosn(φ− ψ)

f(ψ) dψ.

To compute it we represent the sum in the square bracket as a geometric series convergingfor r < ρ:

1

2+∑n≥1

(r

ρ

)ncosn(φ− ψ) =

1

2+ Re

∑n≥1

(r

ρ

)nein(φ−ψ)

=1

2+ Re

r ei(φ−ψ)

ρ− r ei (φ−ψ)=

1

2+

1

2

(r ei(φ−ψ)

ρ− r ei (φ−ψ)+

r e−i(φ−ψ)

ρ− r e−i (φ−ψ)

)

=1

2

ρ2 − r2

ρ2 − 2ρ r cos(φ− ψ) + r2.

In a similar way one can treat the Neumann boundary problem. However in this case onehas to impose an additional constraint for the boundary value of the normal derivative (cf.(4.2.6) above in dimension 1 and (4.2.16)), namely∫

∂Ωg dS = 0. (4.2.28)

We will now prove, for the particular case of a disc domain in the dimension d = 2 that thisnecessary condition of solvability is also sufficient.

Theorem 4.2.6. For an arbitrary C1-smooth 2π-periodic function g(φ) satisfying∫ 2π

0g(φ) dφ = 0 (4.2.29)

the Neumann b.v.p. (4.2.4) in the unitary disc has a solution unique up to an additiveconstant. This solution can be represented by the following integral formula

u(r, φ) =ρ

2π

∫ 2π

0log

ρ2

ρ2 − 2ρ r cos(φ− ψ) + r2g(ψ) dψ. (4.2.30)

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Proof: Repeating the above arguments one arrives at the following expression for the solutionu = u(r, φ):

u =A0

2+∑n≥1

rn (An cosnφ+Bn sinnφ)

(4.2.31)(∂u

∂r

)r=ρ

= g(φ).

Let us consider the Fourier series of the function g(φ)

g(φ) =a0

2+∑n≥1

(an cosnφ+ bn sinnφ) .

Due to the constraint (4.2.29) the constant term vanishes:

a0 = 0.

Comparing this series with the boundary condition (4.2.31) we find that up to an arbitraryconstant

u(r, φ) =∑n≥1

ρ

n

(r

ρ

)n(an cosnφ+ bn sinnφ) ,

where

an =1

π

∫ 2π

0cosnψ g(ψ) dψ, bn =

1

π

∫ 2π

0sinnψ g(ψ) dψ.

Combining these equations we arrive at the following expression:

u(r, φ) =ρ

π

∫ 2π

0

∑n≥1

1

n

(r

ρ

)ncosn(φ− ψ)g(ψ) dψ. (4.2.32)

It remains to compute the sum of this trigonometric series. The stated result (4.2.30) followsby substituting θ = φ− ψ, R = r/ρ in the following Lemma.

Lemma 4.2.7. Let R and θ be two real numbers, R < 1. Then

∞∑n=1

1

nRn cosnθ =

1

2log

1

1− 2R cos θ +R2. (4.2.33)

Proof: The series under consideration can be represented as the real part of a complex series

∞∑n=1

1

nRn cosnθ = Re

∞∑n=1

1

nRneinθ.

The latter can be written as follows:∞∑n=1

1

nRneinθ =

∫ R

0

∞∑n=1

1

RRneinθ dR.

We can easily compute the sum of the geometric series with the denominator Reiθ. Integratingwe obtain

∞∑n=1

1

nRneinθ =

∫ R

0

eiθ

1−ReiθdR = − log

(1−Reiθ

).

Hence∞∑n=1

1

nRn cosnθ =

1

2

[log

1

1−Reiθ+ log

1

1−Re−iθ

]=

1

2log

1

1− 2R cos θ +R2.

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4.3 Properties of harmonic functions: mean value theorem, the maximumprinciple

In this section we will establish two fundamental properties of harmonic functions.

Let Ω ⊂ Rd be a domain. Let x0 ∈ Ω be an internal point and thus there exists someradius R > 0 such that the ball B(x0, R) with the centre at x0 entirely belongs to Ω. Theradius is chosen small enough to guarantee that the sphere belongs to the domain Ω. Weshall work, as in Section 3.8 with the mean value of a continuous function f(x) on the sphereS(x0, R)

f =

∫−S(x0,R)

f(x) dS :=1

dαdRd−1

∫S(x0,R)

f(x) dS. (4.3.1)

In the particular case of a constant function the mean value coincides with the value of thefunction.

Recal that in dimension d = 1 the zero-dimensional “sphere” consists of two points x0±Rand its area is α1 = π1/2/Γ

(32

)= 2. So the mean value of a function is just the arithmetic

mean

f =f(x0 +R) + f(x0 −R)

2.

In the next case d = 2 the sphere is just a circle of radius R with the centre at x0. The area(i.e., the length) element is dS = Rdφ and the mean value is given by (α2 = π)

f =1

2π

∫ 2π

0f(φ) dφ.

Theorem 4.3.1. Let u = u(x) be a function harmonic in Ω. Then the mean value u of uover the sphere S(x0, R) centered at a point x0 ∈ Ω is equal to the value of the function atthis point:

u(x0) =

∫−S(x0,R)

u(x) dS. (4.3.2)

Proof: We give the proof only in the specific case of dimension d = 2. Denote f(φ) therestriction of the harmonic function u onto the small circle |x − x0| = R. By definition thefunction u(x) satisfies the Dirichlet b.v.p. inside the circle:

∆u(x) = 0, |x− x0| < R

u(x)||x−x0|=R = f(φ).

As we already know from the proof of Theorem 4.2.5 the solution to this b.v.p. can berepresented by the Fourier series

u(r, φ) =a0

2+∑n≥1

( rR

)n(an cosnφ+ bn sinnφ) (4.3.3)

for r := |x− x0| < R (cf. (4.2.27) above). In this formula

a0

2=

1

2π

∫ 2π

0f(φ) dφ

is the mean value of the function u on the circle, ie. the r.h.s. of (4.3.3). On the other sidethe value of the function u at the center of the circle can be evaluated substituting r = 0 inthe formula (4.3.3):

u(x0) =a0

2.

Comparing the last two equations we arrive at (4.3.2).

55


Using the mean value theorem we will now prove another important property of harmonicfunctions, namely the maximum principle. Recall that a function u(x) defined on a domainΩ ⊂ Rd is said to have a local maximum at the point x0 if the inequality

u(x) ≤ u(x0) (4.3.4)

holds true for any x ∈ Ω sufficiently close to x0. A local minimum is defined in a similar way.First the following Main Lemma:

Lemma 4.3.2. Let the harmonic function u(x) have a local maximum, or local minimum,at an internal point x0 ∈ Ω. Then u(x) ≡ u(x0) on some neighborhood of the point x0.

Proof: Let us consider the case of a local maximum. Choosing a sufficiently small sphereS(x0, R) with the centre at x0 we can assume the inequality (4.3.4) holds true for all x ∈S(x0, R). Then according to the mean value theorem

u(x0) =

∫−|x−x0|=R

u(x) dS ≤∫−|x−x0|=R

u(x0) dS = u(x0).

If there exists a point x arbitrarily close to x0 such that u(x) < u(x0) then also the inequalityis strict (the smaller value can not be compensated by other points). This is a contradiction,which shows that the function u(x) takes constant values on some ball with the centre at x0.The case of a local minimum can be treated in a similar way.

Theorem 4.3.3. (Maximum principle).Let a function u(x) be harmonic in a bounded connected domain Ω and continuous in a closeddomain Ω. Denote

M = supx∈∂Ω

u(x), m = infx∈∂Ω

u(x).

Then

1) m ≤ u(x) ≤M for all x ∈ Ω;

2) if u(x) = M or u(x) = m for some internal point x ∈ Ω then the function u is constant.

Proof: DenoteM ′ = sup

x∈Ω

u(x)

the maximum of the function u continuous on the compact Ω; it is achieved on some x0,i.e. u(x0) = M ′. We want to prove that M ′ ≤ M . Indeed, let us assume the contrary thatM ′ > M . Then x0 can not belong to ∂Ω but must be internal, x0 ∈ Ω. Denote Ω′ ⊂ Ω theset of points x of the domain where the function u takes the same value M ′. According tothe Main Lemma this subset is open. Clearly it is also closed and nonempty. Hence Ω′ = Ωsince the domain is connected. In other words the function u is constant everywhere in Ω.Because of the continuity it assumes the same value M ′ also at the points of the boundary∂Ω. But there we have M ′ ≤M , which contradicts our assumption.Therefore the value of a harmonic function at an internal point of the domain cannot bebigger than the maximal value of this function on the boundary of the domain. Moreover ifthe harmonic function takes the value M at an internal point then it is constant.In a similar way we prove that a non-constant harmonic function cannot have a minimum inΩ.

Corollary 4.3.4. Given two functions u1(x), u2(x) harmonic in a bounded domain Ω andcontinuous in the closed domain Ω. If

|u1(x)− u2(x)| ≤ ε for x ∈ ∂Ω

56


then|u1(x)− u2(x)| ≤ ε for any x ∈ Ω .

Proof: Denoteu(x) = u1(x)− u2(x).

The function u is harmonic in Ω and continuous in Ω. By assumption we have −ε ≤ u(x) ≤ εfor any x ∈ ∂Ω. So

−ε ≤ infx∈∂Ω

u(x), supx∈∂Ω

u(x) ≤ ε.

According to the maximum principle we must also have

−ε ≤ infx∈Ω

u(x), supx∈Ω

u(x) ≤ ε.

The Corollary implies that the the solution to the Dirichlet boundary value problem, ifexists, depends continuously on the boundary data.

4.4 Harmonic functions on the plane and complex analysis

Recall that a differentiable complex valued function f(x, y) on a domain in R2 is calledholomorphic if its real u(x, y) and imaginary v(x, y) parts satisfy the system of Cauchy –Riemann equations

∂u∂x −

∂v∂y = 0

∂v∂x + ∂u

∂y = 0

(4.4.1)

or, in the complex form∂f

∂x+ i

∂f

∂y= 0.

Introducing complex combinations of the Euclidean coordinates

z = x+ iy, z = x− iy

with inverse

x =z + z

2, y =

z − z2i

(4.4.2)

we have

∂

∂z=

1

2

(∂

∂x− i ∂

∂y

)(4.4.3)

∂

∂z=

1

2

(∂

∂x+ i

∂

∂y

).

So the Cauchy – Riemann equations can be rewritten in the form

∂f

∂z= 0. (4.4.4)

Example. Let f(x, y) be a polynomial

f(x, y) =∑k,l

aklxkyl.

57


It is a holomorphic function iff, after the substitution (4.4.2) there will be no dependence onz (no powers of z): ∑

k,l

akl

(z + z

2

)k (z − z2i

)l=∑m

cmzm.

In that case the result will be a polynomial in z. For example a quadratic polynomial

f(x, y) = ax2 + 2bxy + cy2

is holomorphic iff a+c = 0 (from the coefficient of z/2) and b = i2(a−c) (from the coefficient

of z2/4).

Holomorphic functions are usually denoted f = f(z). The partial derivative ∂/∂z of aholomorphic function is denoted df/dz or f ′(z). One also defines antiholomorphic functionsf = f(z) as differentiable complex functions satisfying the equation

∂f

∂z= 0. (4.4.5)

Notice that the complex conjugate f(z) to a holomorphic function is an antiholomorphicfunction.

From complex analysis it is known that any function f holomorphic on a neighborhood ofa point z0 is also a complex analytic function, i.e., it can be represented as a sum of a powerseries

f(z) =∞∑n=0

an(z − z0)n (4.4.6)

convergent uniformly and absolutely for sufficiently small |z− z0|. In particular it is continu-ously differentiable any number of times. Its real and imaginary parts u(x, y) and v(x, y) areinfinitely smooth functions of x and y.

Theorem 4.4.1. The real and imaginary parts of a function holomorphic in a domain Ω areharmonic functions on the same domain.

Proof: Differentiating the first equation in (4.4.1) in x and the second one in y and addingwe obtain

∂2u

∂x2+∂2u

∂y2= 0.

Similarly, differentiating the second equation in x and subtracting the first one differentiatedin y gives

∂2v

∂x2+∂2v

∂y2= 0.

Corollary 4.4.2. For any integer n ≥ 1 the functions

Re zn and Im zn (4.4.7)

are polynomial harmonic functions (real polynomial solutions to the Laplace equation).

Observe that these polynomials can be represented in the polar coordinates r, φ as

Re zn = rn cosnφ, Im zn = rn sinnφ.

These are exactly the functions we used to solve the main boundary value problems for thedisc.

58


Polynomial harmonic functions are just called harmonic polynomials. We obtain a se-quence of harmonic polynomials

x, y, x2 − y2, xy, x3 − 3xy2, 3x2y − y3, . . . .

Remark Note that the Laplace operator ∆ = ∂2

∂x2+ ∂2

∂y2in the coordinates z, z becomes

∆ = 4∂2

∂z∂z. (4.4.8)

(use (4.4.3)). Using this representation of the two-dimensional Laplace operator one candescribe all complex valued solutions to the Laplace equation.

Theorem 4.4.3. Any complex valued solution u to the Laplace equation ∆u = 0 on the planecan be represented as a sum of a holomorphic and an antiholomorphic function:

u(x, y) = f(z) + g(z). (4.4.9)

Proof: Let the C2-smooth function u(x, y) satisfy the Laplace equation

∂2u

∂z∂z= 0.

Denote

F =∂u

∂z.

The Laplace equation implies that this function is holomorphic, F = F (z). From complexanalysis it is known that any holomorphic function admits a holomorphic primitive f(z), i.e.

F (z) = f ′(z).

Consider the difference g := u− f . It is an antiholomorphic function, g = g(z). Indeed,

∂g

∂z=∂u

∂z− f ′ = 0.

Thusu = f(z) + g(z).

Now the condition u = u that u given by (4.4.9) is real valued, so harmonic, means that

f(z)− g(z) = −g(z) + f(z).

Since the right hand side is holomorphic and the left hand side is its complex conjugate (soantiholomorphic), both must be just a constant real function λ. Hence

g(z) = f(z)− λ

and sou = f(z) + f(z)− λ .

Hence

Corollary 4.4.4. Any harmonic function on the plane can be represented as the real part ofa holomorphic function.

59


Corollary 4.4.5. Any harmonic function on the plane is C∞ (smooth).

If a harmonic u is a polynomial, it is a real part of a holomorphic polynomial. (Iff = v + iw, v must be a polynomial but then by C.R. eqs. also w must be a polynomial).Now a real basis of holomorphic polynomials is zn and izn, and so a real basis of harmonicpolynomials is Rezn and Re izn. Since the imaginary part of a holomorphic function f(z) isequal to the real part of the function −i f(z) (that is holomorphic as well) we have

Corollary 4.4.6. Any harmonic polynomial u is a linear combination of the polynomialsRezn and Im zn as given in (4.4.7).

Another important consequence of the complex representation (4.4.8) of the Laplace op-erator on the plane is invariance of the Laplace equation under conformal transformation.Recall that a smooth map

f : Ω→ Ω′

is called conformal if it preserves the angles between smooth curves (or their tangents). Thedilatations

(x, y) 7→ (k x, k y)

with k 6= 0, rotations by the angle φ

(x, y) 7→ (x cosφ− y sinφ, x sinφ+ y cosφ)

and reflections(x, y) 7→ (x,−y)

are examples of linear conformal transformations. These examples and their superpositionsexhaust the class of linear conformal maps. The general conformal maps on the plane aredescribed by

Lemma 4.4.7. Let f(z) be a function holomorphic in the domain Ω with nowhere vanishingderivative:

df(z)

dz6= 0 ∀z ∈ Ω.

Then the mapz 7→ f(z)

of the domain Ω to Ω′ = f(Ω) is conformal. Same for antiholomorphic functions. Conversely,if the smooth map (x, y) 7→ (u(x, y), v(x, y)) is conformal then the function f = u + iv isholomorphic or antiholomorphic with nonvanishing derivative.

Proof: Let us consider the differential of the map (x, y) 7→ (u(x, y), v(x, y)) given by the realu = Re f and imaginary v = Im f parts of the holomorphic function f . It is a linear mapdefined by the Jacobi matrix ∂u/∂x ∂u/∂y

∂v/∂x ∂v/∂y

=

∂u/∂x −∂v/∂x

∂v/∂x ∂u/∂x

(we have used the Cauchy – Riemann equations). Since

0 6= |f ′(z)|2 =

(∂u

∂x

)2

+

(∂v

∂x

)2

,

we can introduce the numbers r > 0 and φ by

r = |f ′(z)|, cosφ =∂u/∂x√(

∂u∂x

)2+(∂v∂x

)2 , sinφ =∂v/∂x√(

∂u∂x

)2+(∂v∂x

)2 .60


The Jacobi matrix then becomes a combination of the rotation by the angle φ and a dilatationwith the coefficient r: ∂u/∂x ∂u/∂y

∂v/∂x ∂v/∂y

= r

cosφ − sinφ

sinφ cosφ

.

This is a linear conformal transformation preserving the angles. A similar computation worksfor an antiholomorphic map with nonvanishing derivatives f ′(z) 6= 0.

Conversely, the Jacobi matrix of a conformal transformation must have the form

r

cosφ − sinφ

sinφ cosφ

or

r

cosφ sinφ

sinφ − cosφ

.

In the first case one obtains the differential of a holomorphic map while the second matrixcorresponds to the antiholomorphic map.

We are ready to prove

Theorem 4.4.8. Letf : Ω→ Ω′

be a conformal map. Then the pull-back of any function harmonic in Ω′ is harmonic in Ω.

Proof: According to the Lemma the conformal map is given by a holomorphic or an anti-holomorphic function. Let us consider the holomorphic case,

z 7→ w = f(z).

The transformation law of the Laplace operator under such a map is clear from the followingformula:

∂2

∂z∂z= |f ′(z)|2 ∂2

∂w∂w. (4.4.10)

Thus any function U on Ω′ satisfying

∂2U

∂w∂w= 0

will also satisfy∂2U

∂z∂z= 0.

The case of an antiholomorphic map can be considered in a similar way.

A conformal mapf : Ω→ Ω′

is called conformal transformation if it is one-to-one. In that case the inverse map

f−1 : Ω′ → Ω

exists and is also conformal. The following fundamental Riemann mapping theorem is thecentral result of the theory of conformal transformations on the plane.

61


Theorem 4.4.9. For any connected and simply connected domain Ω on the plane not coin-ciding with the plane itself there exists a conformal transformation of Ω to the unit disc.

The Riemann theorem, together with conformal invariance of the Laplace equation givesa possibility to reduce the main boundary value problems for any connected simply connecteddomain to similar problems for the unit disc.

4.5 Exercises to Section 4

Exercise 4.5.1. Find a function u(x, y) satisfying

∆u = x2 − y2

for r < a and the boundary condition u|r=a = 0.

Exercise 4.5.2. Find a harmonic function on the annular domain

a < r < b

with the boundary conditions

u|r=a = 1,

(∂u

∂r

)r=b

= cos2 φ.

Exercise 4.5.3. Find solution u(x, y) to the Dirichlet b.v.p. in the rectangle

0 ≤ x ≤ a, 0 ≤ y ≤ b

satisfying the boundary conditions

u(0, y) = Ay(b− y), u(a, y) = 0

u(x, 0) = B sinπx

a, u(x, b) = 0.

Hint: use separation of variables in Euclidean coordinates.

62


5 Heat equation

5.1 Derivation of heat equation

The heat equation for the temperature u(x, t), as a function of x ∈ Rd and t ∈ R>0, reads

∂u

∂t= a2∆u. (5.1.1)

Here ∆ is the Laplace operator in Rd. We will consider only the case of constant coeffi-cients a = const. For d = 3 this equation describes the distribution of temperature in thehomogeneous and isotropic media at the moment t.

The derivation of heat equation is based on the following assumptions.

1. The heat δQ necessary for changing from u1 to u2 the temperature of a piece of massm is proportional to the mass and to the difference of temperatures:

δQ = cpm(u2 − u1).

The coefficient cp is called specific heat capacity.

2. The Fourier law describing the quantity of heat spreading through a surface S duringthe time interval δt. It says that this quantity δQ is proportional to the area A(S) of thesurface, to the time δt and to the derivative of the temperature u along the normal n to thesurface:

δQ = −k A(S)∂u

∂nδt.

Here the coefficient k > 0 is called thermal conductivity. The negative sign means that theheat is spreading from hot to cold regions.

In order to derive the heat equation let us consider the heat balance within a domainΩ ⊂ Rd with a smooth boundary ∂Ω. The total change of heat in the domain during thetime interval δt is

δQ =

∫Ωcpρ [u(t+ δt, x)− u(t, x)] dV ·

where ρ is the mass density, such that the mass of the media contained in the volume is equalto

m =

∫Ωρ dV.

In the case of a homogeneous media the mass density is constant, so the heat quantity isequal to

δQ ' cpρ∫

Ω

∂u

∂tδt dV.

This change of heat must be equal, with the negative sign, to the one passing through theboundary ∂Ω

δQ =

∫∂Ωk∂u

∂ndS · δt.

Using Green formula we can rewrite this heat flow in the form

k

∫Ω

∆u dV · δt.

Dividing by δt we arrive at the equation

cpρ

∫Ω

∂u

∂tdV = k

∫Ω

∆u dV.

63


Since the domain Ω is arbitrary this gives the heat equation with the coefficient (calledthermal diffusivity)

a2 =k

cpρ.

The heat equation governs heat diffusion, as well as other diffusive processes, such asparticle diffusion or the propagation of action potential in nerve cells. The heat equation isused in probability and describes random walks. For this reason it is also applied in financialmathematics (Black-Scholes or the Ornstein-Uhlenbeck processes). The equation, and variousnon-linear analogues, has also been used in image analysis. It is also important in Riemanniangeometry and surprisingly in topology: it was adapted by R. Hamilton when he defined theRicci flow that was later used by G. Perelman to solve the topological Poincare conjecture.

5.2 Main boundary value problems for heat equation

The simplest is the Cauchy problem of finding a function u(x, t) satisfying

∂u

∂t= a2∆u

(5.2.1)

u(x, 0) = φ(x), x ∈ Rd.

The physical meaning of this problem is clear: given the initial temperature distribution inthe space to determine the temperature at any time t > 0 at any point x of the space.

Often we are interested in the temperature distribution only within the bounded domainΩ ⊂ Rd. In this case one has to add to the Cauchy data within Ω also the information aboutthe temperature on the boundary ∂Ω or about the heat flux through the boundary. In thisway we arrive at two main mixed problems in a bounded domain:

The first mixed problem: find a function u(x, t) satisfying

∂u

∂t= a2∆u, t > 0, x ∈ Ω

u(0, x) = φ(x), x ∈ Ω (5.2.2)

u(x, t) = f(x, t), t > 0, x ∈ ∂Ω.

The second mixed problem is obtained from (5.2.2) by replacing the last condition by(∂u

∂n

)x∈∂Ω

= g(x, t), t > 0, x ∈ ∂Ω, (5.2.3)

where n is the unit external normal to the boundary.

In the particular case of the boundary data independent of time

f = f(x) or g = g(x)

one can look for a stationary solution u satisfying

∂u

∂t= 0.

In this case the first and the second mixed problem for the heat equation reduce respectivelyto the Dirichlet and Neumann boundary value problem for the Laplace equation in Rd.

64


5.3 Fourier transform

Our next goal is to solve the one-dimensional Cauchy problem for heat equation on the line.To this end we will develop a continuous analogue of Fourier series.

Let f(x) be an absolutely integrable complex valued function on the real line, i.e.,∫ ∞−∞|f(x)| dx <∞. (5.3.1)

Definition 5.3.1. The function f ,

f(p) :=1

2π

∫ ∞−∞

f(x)e−ipxdx (5.3.2)

of the real variable p is called the Fourier transform of f(x).

Due to the condition (5.3.1) the integral converges absolutely and uniformly with respectto p ∈ R. Thus the function f(p) is continuous in p.

Example 1. Let us compute the Fourier transform of the Gaussian function

f(x) = e−12x2 .

We have ∫ ∞−∞

e−12x2−ipxdx =

∫ ∞−∞

e−12

(x+ip)2− 12p2dx .

We want to perform a (complex) change of variables

z = x+ ip.

To do this one can consider the integral∮Ce−

12z2− 1

2p2dz, z = x+ iy (5.3.3)

over the boundary C of the rectangle on the complex z-plane

−R ≤ x ≤ R, 0 ≤ y ≤ p.

This integral is equal to zero since the integrand is holomorphic on the entire complex plane.But it is easy to see that the integrals over the vertical segments x = ±R, 0 ≤ y ≤ p in(5.3.3) tend to zero when R→∞ (due to e−R

2and limits on y).

Hence the integrals over the upper and lower horizontal segments∫ −RR

e−12

(x+ip)2− 12p2dx+

∫ R

−Re−

12x2− 1

2p2dx→ 0 as R→∞,

so ∫ ∞−∞

e−12

(x+ip)2− 12p2dx =

∫ ∞−∞

e−12x2− 1

2p2dx .

Using the Euler integral ∫ ∞−∞

e−12x2dx =

√2π (5.3.4)

we finally obtain the Fourier transform of the Gaussian function

f(p) =1√2πe−

12p2 . (5.3.5)

65


Example 2. We compute now the Fourier transform of the function

f(x) =1

x

(which is in L2, but not in L1 at ∞, and at 0 should be considered as a distribution). Wecalculate first the following auxiliary expression

I(a, p) :=

∫ ∞0

e−axsin px

xdx, for a > 0.

We have∂

∂aI(a, p) = −

∫ ∞0

e−axsin px dx = −Im∫ ∞

0e−axeipx dx

= −Im 1

ip− ae(ip−a)x|∞0 = Im

1

a− ip(0− 1) = − p

a2 + p2.

Integrating this from 0 to ∞ we get

I(∞, p)− I(0, p) == −∫ ∞

0

p

a2 + p2da = − arctan

a

p|∞0 = −π

2signp− 0.

But I(∞, p) = 0 and hence I(0, p) = π2 signp. This shows that

f(p) =1

2π

∫ ∞−∞

e−ipx

xdx = − i

π

∫ ∞0

sin px

xdx =: − i

πI(0, p) = − i

2signp .

Corollary 5.3.2. Dirichlet integral:∫ ∞0

sinx

xdx = I(0, 1) =

π

2. (5.3.6)

Lemma 5.3.3 (Riemann–Lebesgue). Let a continuous function f(x) be absolutely integrableon R. Then

limλ→±∞

∫ ∞−∞

f(x)eiλxdx = 0. (5.3.7)

Proof: Because of convergence of the integral∫∞−∞ f(x) dx the difference∫ ∞

−∞f(x) dx−

∫ b

af(x) dx

tends to zero when a→ −∞ b→∞. So it suffices to prove the Lemma for the finite integral.Because of integrability of f(x) there exists, for any given ε > 0, a partition of the interval

a = x0 < x1 < · · · < xn = b

such that

0 <

∫ b

af(x) dx−

n∑j=1

mj δxj < ε, (5.3.8)

whereδxj = xj − xi−1

mj = infx∈[xi−1,xj ]

f(x).

66


Introduce a step-like function

g(x) = mj for x ∈ [xi−1, xj ], i = 1, . . . , n.

Then ∣∣∣∣∫ b

af(x)eiλxdx−

∫ b

ag(x)eiλxdx

∣∣∣∣ ≤ ∫ b

a|f(x)− g(x)|

∣∣∣eiλx∣∣∣ dx=

∫ b

a|f(x)− g(x)| dx < ε

by (5.3.8). But the integral∫ b

ag(x)eiλxdx =

n∑j=1

mj

i λ

(eiλxj − eiλxj−1

)tends to zero when λ→ ±∞ and so also the integral

∫ ba g(x)eiλxdx . This shows (5.3.7).

We will now establish, under certain additional assumptions, validity of the inversionformula for the Fourier transform:∫ ∞

−∞f(p)eipxdp = f(x). (5.3.9)

Theorem 5.3.4. Let the absolutely integrable function f(x) be differentiable at any pointx ∈ R. Then

limR→∞

∫ R

−Rf(p)eipxdp = f(x). (5.3.10)

Proof: Denote IR(x) the integral in the left hand side of (5.3.10). Using continuity anduniform convergence of the Fourier integral (5.3.2) we can apply Fubini theorem to thisintegral and thus rewrite it as follows:

IR(x) =

∫ R

−Rf(p)eipxdp =

∫ R

−R

(1

2π

∫ ∞−∞

f(y)e−ipydy

)eipxdp

=1

2π

∫ ∞−∞

f(y)

(∫ R

−Reip(x−y)dp

)dy =

1

π

∫ ∞−∞

f(y)sinR(x− y)

x− ydy

=1

π

∫ ∞−∞

f(x+ s)sinRs

sds =

1

π

∫ ∞0

[f(x+ s) + f(x− s)] sinRs

sds.

Using the Dirichlet integral (5.3.6) we can rewrite the difference of the l.h.s. and r.h.s. in theform

IR(x)− f(x) =1

π

∫ ∞0

f(x+ s)− 2f(x) + f(x− s)s

sinRsds.

Because of the differentiablity requirement on f we have

lims→0

f(x+ s)− 2f(x) + f(x− s)s

= lims→0

f(x+ s)− f+(x)

s+ lims→0

f(x− s)− f−(x)

s

= f ′+(x)− f−′(x) = 0,

and so the part

F (s;x) =

f(x+s)−2f(x)+f(x−s)

s , s 6= 0

0, s = 0

67


of the integrand is a continuous function in s depending on the parameter x.

Now represent the last integral in the form∫ ∞0

f(x+ s)− 2f(x) + f(x− s)s

sinRsds

=

∫ 1

0F (s;x) sinRsds+

∫ ∞1

f(x+ s) + f(x− s)s

sinRsds− 2f(x)

∫ ∞1

sinRs

sds.

Now in the limit R → ∞ the first integral in the r.h.s. vanishes according to the proof ofRiemann–Lebesgue lemma. The same is true for the second integral (there is no pole fors ∈ [1,∞]). Finally the last integral by a change of integration variable x = Rs reduces to∫ ∞

1

sinRs

sds =

∫ ∞R

sinx

xdx→ 0 for R→∞

(since the Dirichlet integral converges).

Proposition 5.3.5. Let f(x) be an absolutely integrable piecewise continuous function of x ∈R differentiable on every interval of continuity. Let us also assume that at every discontinuitypoint x0 the left and right limits f−(x0) and f+(x0) exists and, moreover, the left and rightderivatives

lims→0−

f(x0 + s)− f−(x0)

sand lim

s→0+

f(x0 + s)− f+(x0)

s

exist as well. Then there is following modification of the inversion formula for the Fouriertransform

limR→∞

∫ R

−Rf(p)eipxdp =

f(x), x is a continuity point

f−(x)+f+(x)2 , x is a discontinuity point

(5.3.11)

Proof: Vedere le modifiche in rosso e sostituire f = f+(x)+f−(x)2 and 0 = f ′+(x)− f ′−(x) nella

prova precedente.

The main property of Fourier transform used for solving linear PDEs is given by thefollowing formula:

Lemma 5.3.6. Let f(x) be an absolutely integrable continuously differentiable function withabsolutely integrable derivative f ′(x). Then∫ ∞

−∞f ′(x)e−ipxdx = ip f(p). (5.3.12)

Proof: From the integrability of f ′(x) it follows that there exist limits

f(±∞) := limx→±∞

f(x) = f(0) + limx→±∞

∫ x

0f ′(y) dy.

Because of absolute integrability of f the limiting values f(±∞) must be equal to zero.Integrating by parts∫ ∞

−∞f ′(x)e−ipxdx =

(e−ipxf(x)

)∞−∞ + ip

∫ ∞−∞

f(x)e−ipxdx = ipf(p),

but the first term on the r.h.s. vanishes and we arrive at the needed formula.

68


Denote F the linear map of the space of functions in x variable to the space of functionsin p variable given by the Fourier transform:

F(f) = f . (5.3.13)

The property formulated in the above Lemma says that the operator of x-derivative trans-forms to i times the operator of multiplication by the independent variable

d

dxf (p) = ip f(p). (5.3.14)

This property of the Fourier transform will be used in the next section for solving the Cauchyproblem for heat equation.

A similar calculation gives the formula

xf(p) = id

dpf(p) (5.3.15)

valid for functions f = f(x) absolutely integrable together with xf(x). Moreover, the productof two functions transforms to the convolution product of their transforms

fg(p) = 2π(f ∗ g

)(p) := 2π

∫f(q)g(p− q) dq . (5.3.16)

We leave the proof of this formulae as an exercise for the reader. Finally we remark that theFourier transform in Rd is defined as the operation of Fourier transform in each coordinateR.

5.4 Solution to the Cauchy problem for heat equation on the line

Let us consider the one-dimensional Cauchy problem for the heat equation

∂u

∂t= a2∂

2u

∂x2, t > 0

(5.4.1)

u(x, 0) = φ(x), x ∈ R.

Theorem 5.4.1. Let the initial data φ(x) be absolutely integrable function on R. Then theCauchy problem (5.4.1) has a unique solution u(x, t) absolutely integrable in x ∈ R for allt > 0 represented by the formula

u(x, t) =

∫ ∞−∞

G(x− y; t)φ(y) dy. (5.4.2)

where

G(x; t) =1

2a√πte−

x2

4a2t . (5.4.3)

The integral representation (5.4.2) of solutions to the Cauchy problem is called Poissonintegral and its integral kernel is called heat kernelThere are similar expressions in Rd, one

has to change the prefactor to (4πt)−d2 a−1 and x to |x|.

Proof: Denote

u(p, t) =1

2π

∫ ∞−∞

u(x, t)e−ipxdx

69


the Fourier-image of the unknown solution. According to Lemma 5.3.6 the function u(p, t)satisfies equation

∂u(p, t)

∂t= −a2p2u(p, t).

This equation can be easily solved

u(p, t) = u(p, 0)e−a2p2t.

Due to the initial condition we obtain

u(p, 0) = φ(p) =1

2π

∫ ∞−∞

φ(y)e−ipydy.

Thusu(p, t) = φ(p)e−a

2p2t. (5.4.4)

It remains to apply the inverse Fourier transform to this formula:

u(x, t) =

∫ ∞−∞

eixpφ(p)e−a2p2tdp =

1

2π

∫ ∞−∞

(eixp−a

2p2t

∫ ∞−∞

e−ipyφ(y) dy

)dp

=1

2π

∫ ∞−∞

φ(y)

(∫ ∞−∞

eip(x−y)−a2p2tdp

)dy

The integral in p is nothing but the (inverse) Fourier transform of the Gaussian function. Acalculation similar to Example 1 gives the value for this integral∫ ∞

−∞eip(x−y)−a2p2tdp =

√π

a√te−

(x−y)2

4a2t

(change the variables p = q/√

2ta). This completes the proof of the Theorem.

Remark 5.4.2. The formula (5.4.2) can work also for not necessarily absolutely integrablefunctions. For example for the constant initial data φ(x) ≡ φ0 we obtain u(x, t) ≡ φ0 due tothe following integral

1

2a√πt

∫ ∞−∞

e−(x−y)2

4a2t dy ≡ 1. (5.4.5)

We will now use this integral and the Poisson integral (5.4.2) in order to prove an analogueof the maximum principle for solutions to the heat equation.

Theorem 5.4.3. Let m = infx∈R φ(x) and M = supx∈R φ(x). The solution to the Cauchyproblem represented by the Poisson integral (5.4.2) for all t > 0 satisfies

m ≤ u(x, t) ≤M. (5.4.6)

Moreover, if some of the inequalities becomes equality for some t > 0 and x ∈ R then u(x, t) ≡const.

Proof: Multiply m ≤ φ(y) ≤M by a (positive) G(x− y; t), integrate over y and use (5.4.5).The equality u(x, t) = M for some x, t means that

∫G(x− y; t)(M −φ(y))dy = 0, which due

to the same positivity can have place only if φ(x) ≡ M but then also u(x, t) ≡ M . Similarreasoning applies if u(x, t) = m.

Corollary 5.4.4. The solution to the Cauchy problem (5.4.1) for the heat equation dependscontinuously on the initial data.

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Proof: Let u1(x, t), u2(x, t) be two solutions to the heat equation with the initial data φ1(x)and φ2(x) respectively. If the initial data differ by ε, i.e.

|φ1(x)− φ2(x)| ≤ ε ∀x ∈ R

then from the maximum principle applied to the solution u(x, t) = u1(x, t)−u2(x, t) it followsthat

|u1(x, t)− u2(x, t)| ≤ ε.

5.5 Mixed boundary value problems for the heat equation

Let us begin with the 2π-periodic problem

∂u

∂t= a2∂

2u

∂x2, t > 0,

u(x+ 2π, t) = u(x, t), t > 0, (5.5.1)

u(x, 0) = φ(x),

where φ(x) is a smooth 2π-periodic function.

Theorem 5.5.1. There exists a unique solution C2 in x and C1 in t to the problem (5.5.1).It can be represented in the form

u(x, t) =1

2π

∫ 2π

0Θ(x− y; t)φ(y) dy, t > 0 (5.5.2)

where

Θ(x; t) =∑n∈Z

e−a2n2t+inx. (5.5.3)

Proof: Let us expand the unknown periodic function u(x, t) in the Fourier series:

u(x, t) =∑n∈Z

un(t)einx,

where

un(t) =1

2π

∫ 2π

0u(x, t)e−inxdx.

The substitution to the heat equation yields

∂un(t)

∂t= −a2n2un(t),

soun(t) = un(0)e−a

2n2t, n ∈ Z.

At t = 0 one must meet the initial conditions, so

un(0) = φn =1

2π

∫ 2π

0φ(y)e−inydy.

For the function u(x, t) we obtain

u(x, t) =1

2π

∑n∈Z

∫ 2π

0e−a

2n2t+in(x−y)φ(y)dy.

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In order to show (5.5.2) it suffices to exchange the sum with the integral, which is possiblesince the series (5.5.3) converges absolutely and uniformly for all x ∈ R and all t > ε for anyε > 0. This easily follows from the uniform convergence of the integral (which superates theseries Θ− 1) ∫ ∞

0e−a

2y2tdy =1

a√t

∫ ∞0

e−y2dy <∞ for t > ε > 0.

So far we have uniqueness: if the solution exists, it must have the form (5.5.2). In a similarway one can prove that the series (5.5.3) can be differentiated any number of times (the seriesof derivatives converges uniformly). Moreover (5.5.2) is C1 in t and C2 in x (we can deriveunder the integration sign) and so it is indeed a solution. The theorem is proved.

We shall dedicate now some time to see the relations with other mathematical structures.The functioned defined by the series (5.5.3) is called theta-function. It is expressed via theJacobi theta-function

θ3(φ | τ) =∑n∈Z

eπin2τ+2πinφ (5.5.4)

by a change of variables

Θ(x; t) = θ(φ | τ), φ =1

2πx, τ = i

a2t

π. (5.5.5)

The convergence of the series (5.5.4) for Jacobi theta function takes place for all complexvalues of τ provided

Im τ > 0 (t > 0). (5.5.6)

The function Θ(x; t) is periodic in x with the period 2π while the Jacobi theta-function isperiodic in φ with the period 1. It satisfies many remarkable properties. Let us see some ofthem.

Lemma 5.5.2. Θ(x; t) is a real valued function on R× R+ and∫ 2π

0Θ(x; t) dx = 2π. (5.5.7)

Proof: Change of the summation variable n→ −n in (5.5.3) shows that

Θ(x; t) = Θ(x; t).

Exchanging the integral and sum,∑∫e−n

2t+inx dx =∑

2πδn,0 = 2π .

Lemma 5.5.3. The series ∑n∈Z

e−a2n2t+inz (5.5.8)

converges for any complex number z = x+ iy uniformly on the strips |Im z| = y ≤M for anypositive M . So the theta-function (5.5.3) can be analytically continued to a function Θ(z; t)holomorphic on the entire complex z-plane.

Proof: Adapt the proof of theorem 5.5.1

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Lemma 5.5.4. The function Θ(z; t) is even and 2π-periodic in z, and satisfies the identity

Θ(z + 2ia2t; t

)= ea

2t−izΘ(z; t). (5.5.9)

The complex number 2ia2t is called quasi-period of the theta-function.

Proof: Since z 7→ −z together with n 7→ −n leaves invariant (5.5.3), Θ is even in z. Theperiodicity under z 7→ z + 2π is evident. Moreover,∑

n∈Ze−a

2n2t+in(z+2ia2t) =∑n∈Z

e−a2t(n2+2n+1−1)ei(n+1−1)z = ea

2t−iz∑n∈Z

e−a2(n+1)2t+i(n+1)z

shows the quasi-periodicity.

Lemma 5.5.5. The theta-function has zeroes at the points

z = zk l = π(2k + 1) + ia2t (2l + 1), k, l ∈ Z. (5.5.10)

Proof: Due to the 2π-periodicity and 2ia2t quasi-periodicity it suffices to check k = 0 = l.Set ω = π + ia2t. We have

Θ(−z + ω) = Θ(−z − ω + 2π + 2ia2t) = Θ(−z − ω)ea2t+i(z+ω) = Θ(z + ω)(−eiz),

where in the last equality we used the parity of Θ and eiπ = −1. Setting z = 0 we get thestatement

Θ(ω) = −Θ(ω) = 0.

Lemma 5.5.6. The theta-function has no other zeroes on the complex plane. Moreover

Θ(x; t) > 0 for x ∈ R. (5.5.11)

Proof: If z = zi, i = 1, . . . , n are zeroes of a function f holomorphic in (bounded) Ω, thenF (z) = f(z)(z − z1) . . . (z − zn), where f is holomorphic and has no zeroes in Ω, and

1

2πi

∮∂Ω

dF (z)

F (z)=

1

2πi

∮∂Ω

∑i

dz

z − zi= n.

Now, this integral for F (z) = Θ(z; t) (parametric in t) and Ω given as the rectangle 0 ≤x ≤ 2π, 0 ≤ y ≤ 2a2t on the complex z-plane, z = x+ iy, gives the result 1. This comes asfollows. From the 2π-periodicity of Θ the contributions of the oriented vertical segments of∂Ω cancel. From the quasi-periodicity of Θ(z; t), the contributions of the oriented horizontalsegments cancel, except (due to the Leibniz rule) the contribution of the upper segment

1

2πi

∫ 0

2π

d e−iz

e−iz= (−)(−)

i

2πi

∫ 2π

0dx = 1.

Since Θ(x; t) is real, has no zeroes on R, and Θ(0; t) > 0, it must be positive.

Another proof of positivity of the theta-function follows from the following Poisson sum-mation formula that is of course of interest on its own.

Lemma 5.5.7. Let f(x) be a continuously differentiable absolutely integrable function satis-fying the inequalities

|f(x)| < C(1 + |x|)−1−ε, |f(p)| < C(1 + |p|)−1−ε

for some positive C and ε. Here f(p) is the Fourier transform of f(x). Then∑n∈Z

f(2πn) =∑m∈Z

f(m). (5.5.12)

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Proof: We will actually prove a somewhat more general formula∑n∈Z

f(x+ 2πn) =∑m∈Z

f(m)eimx, x ∈ R. (5.5.13)

Of course since f is not necessarily periodic, the right hand side need not to be its Fourierseries and we have to work out various properties anew. The series on the l.h.s. and the serieson the r.h.s. (as well as the series of derivatives of their terms) converge uniformly due tothe decay conditions on f and f . Thus on both sides we have differentiable functions whichare moreover 2π-periodic in x. Therefore, it suffices to check that the Fourier coefficients cmof the function on the left side coincide with the Fourier coefficients of the function on theright side, which are clearly f(m). Indeed, the m-th Fourier coefficient of the left hand sideis equal to

cm =1

2π

∫ 2π

0

(∑n∈Z

f(x+ 2πn)

)e−imxdx.

Due to absolute and uniform (in x) convergence of the series∑n∈Z

f(x+ 2πn)

one interchange the order of summation and integration to arrive at

cm =1

2π

∑n∈Z

∫ 2π

0f(x+ 2πn)e−imxdx.

By shifting in the n-th summand the integration variable

y = x+ 2πn,

(and accordingly shifting the integration limits) one rewrites the sum as follows:

cm =1

2π

∑n∈Z

∫ 2π(n+1)

2πnf(y)e−imy−2πimndy =

1

2π

∫ ∞−∞

f(y)e−imydy = f(m),

since e−2πimn = 1.

Using the Poisson summation formula we can prove the following remarkable identity forthe theta-function.

Proposition 5.5.8. The theta-function (5.5.1) satisfies the following identity

Θ(x; t) =1

a

√π

t

∑n∈Z

e−(x+2πn)2

4a2t . (5.5.14)

Proof: It can be obtained by applying the Poisson summation formula (in the strongerversion (5.5.13)) to the function

f(x) =1

a

√π

te−

x2

4a2t

which gives the r.h.s. of (5.5.14). Its Fourier transform

f(p) = e−a2p2t,

we computed in Theorem 5.4.1, and this yields the defining formula for (5.5.1). The requiredassumptions on f and f are clearly satisfied.

Now, looking at (5.5.14), we can affirm that Θ(x; t) is positive.

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Remark 5.5.9. The formula (5.5.14) is the clue to derivation of the transformation law forthe Jacobi theta-function under modular transformations

τ 7→ aτ + b

cτ + d, a, b, c, d ∈ Z, ad− bc = 1.

Let us now consider the first mixed problem for heat equation on the interval [0, l] withzero boundary conditions:

∂u

∂t= a2∂

2u

∂x2, 0 ≤ x ≤ l, t > 0

u(0, t) = u(l, t) = 0 (5.5.15)

u(x, 0) = φ(x), 0 ≤ x ≤ l.

Analogously to Section 3.6 let us extend the initial data φ(x) to the real line as an odd2l-periodic function. We leave as an exercise for the reader to check that the solution to thisperiodic Cauchy problem will remain an odd periodic function for all times and, hence, itwill vanish at the points x = 0 and x = l. In this way one arrives at the following

Theorem 5.5.10. The mixed b.v.p. (5.5.15) has a unique solution for an arbitrary smoothfunction φ(x). It can be represented by the following integral

u(x, t) =1

l

∫ l

0Θ(x, y; t)φ(y) dy (5.5.16)

where

Θ(x, y; t) = 2∞∑n=1

e−a2n2t sin

πnx

lsin

πny

l. (5.5.17)

5.6 More general boundary conditions for the heat equation. Solution tothe inhomogeneous heat equation

In the previous section the simplest b.v.p. for the heat equation has been considered. Wewill now address the more general problem

∂u

∂t= a2∂

2u

∂x2, t > 0, 0 < x < l (5.6.1)

u(0, t) = f0(t), u(l, t) = f1(t), t > 0

u(x, 0) = φ(x), 0 < x < l.

The following simple procedure reduces the above problem for u(x, t) to the b.v.p. for v(x, t)with zero boundary condition for the inhomogeneous heat equation in d = 1. Namely, letu = v + w, where

w(x, t) :=[f0(t) +

x

l(f1(t)− f0(t))

]. (5.6.2)

(Note for the next case that w is a (unique) harmonic in d = 1). Note also that w(x, t) takesthe needed values f0(t) and f1(t) at the endpoints of the interval. Now, v = u − w indeedsatisfies

∂v

∂t= a2 ∂

2v

∂x2+ F (x, t), t > 0, 0 < x < l (5.6.3)

v(0, t) = v(l, t) = 0, t > 0

v(x, 0) = Φ(x), 0 < x < l ,

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where the functions F (x, t), Φ(x) are given by

F (x, t) = −∂w(x, t)

∂t= −

[df0(t)

dt+x

l

(df1(t)

dt− df0(t)

dt

)](5.6.4)

Φ(x) = φ(x)− w(x, 0) = φ(x)−[f0(0) +

x

l(f1(0)− f0(0))

].

In the more general case of d-dimensional heat equation with non-vanishing boundaryconditions

∂u

∂t= a2∆u, t > 0, x ∈ Ω ⊂ Rd (5.6.5)

u(x, t)|x∈∂Ω = f(x, t), t > 0

u(x, 0) = φ(x), x ∈ Ω

the procedure is similar to the above one. Namely, denote w(x, t) the solution to the Dirichletboundary value problem for the Laplace equation in x depending on t as on the parameter:

∆w = 0, x ∈ Ω ⊂ Rd (5.6.6)

w(x, t)|x∈∂Ω = f(x, t).

We already know that the solution to the Dirichlet boundary value problem is unique anddepends continuously on the boundary conditions. Therefore the solution w(x, t) is a contin-uous function on Ω×R>0. One can also prove that this functions is smooth, if the boundarydata f(x, t) are so. Then the substitution

u(x, t) = v(x, t) + w(x, t) (5.6.7)

reduces the mixed b.v.p. (5.6.5) to the one with zero boundary conditions

v(x, t)|x∈∂Ω = 0, t > 0

with the modified initial data

v(x, 0) = φ(x)− w(x, 0), x ∈ Ω

but the heat equation becomes inhomogeneous one:

∂v

∂t= a2∆v + F (x, t), F (x, t) = −∂w(x, t)

∂t, x ∈ Ω.

We will now explain a simple method for solving the inhomogeneous heat equation. Forthe sake of simplicity let us consider in details the case of one spatial variable. Moreover wewill concentrate on the infinite line case. So the problem under consideration is in finding afunction u(x, t) on R× R>0 satisfying

∂u

∂t= a2∂

2u

∂x2+ f(x, t), x ∈ R, t > 0 (5.6.8)

u(x, 0) = φ(x).

Theorem 5.6.1. The solution to the inhomogeneous problem (5.6.8) has the form

u(x, t) =

∫ t

0ds

∫ ∞−∞

G(x− y; t− s) f(y, s)dy +

∫ ∞−∞

G(x− y; t)φ(y) dy (5.6.9)

where the function G(x; t) was defined in (5.4.3).

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Proof: The first term

u1(x, t) =

∫ t

0ds

∫ ∞−∞

G(x− y; t− s) f(y, s)dy

clearly vanishes at t = 0. Let us prove that it satisfies the inhomogeneous heat equation

∂u1

∂t= a2∂

2u1

∂x2+ f(x, t).

Denote

v(x, t; s) =

∫ ∞−∞

G(x− y; t− s) f(y, s)dy.

Like in the Theorem 5.4.1 we derive that this is a solution to the homogeneous heat equationin x, t depending on the parameter s. This solution is defined for t ≥ s; for t = s it satisfiesthe initial condition vt in 2.9 but there we had ∂2

tv(x, s; s) = f(x, s).

Now applying ∂∂t to the first term

u1(x, t) =

∫ t

0v(x, t; s) ds

one obtains

∂

∂tu1(x, t) = v(x, t; t) +

∫ t

0

∂

∂tv(x, t; s) ds = f(x, t) +

∫ t

0a2 ∂

2

∂x2v(x, t; s) ds

= f(x, t) + a2 ∂2

∂x2u1(x, t) .

Concerning the the second term

u2(x, t) =

∫ ∞−∞

G(x− y; t)φ(y) dy

in (5.6.9), we already know from Theorem 5.4.1 that it solves the homogeneous heat equationand satisfies initial condition

u2(x, 0) = φ(x).

This concludes the proof.

Ricapitoliamo alcune proprieta salienti dell’equazione del calore in R2:- superposizioni/scomposizioni di soluzioni (in ’modi’ semplici)- esistenza e unicita di soluzioni e problema di Cauchy ben posto- propagazione istantanea- smorzatura di singolarita nel tempo (per il principio del massimo)- nel caso periodico legame con funzioni theta.

5.7 Exercises to Section 5

Exercise 5.7.1. Let the function f(x) belong to the class Ck(R) and, moreover, all thefunctions f(x), f ′(x), . . . , f (k)(x) be absolutely integrable on R. Prove that then

f(p) = O(

1

pk

)for |p| → ∞. (5.7.1)

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Exercise 5.7.2. Let f(p) be the Fourier transform of the function f(x). Prove that eiapf(p)is the Fourier transform of the shifted function f(x+ a).

Exercise 5.7.3. Find Fourier transforms of the following functions.

f(x) = ΠA(x) =

1

2A , |x| < A0, otherwise

(5.7.2)

f(x) = ΠA(x) cosωx (5.7.3)

f(x) =

1A

(1− |x|A

), |x| < A

0, otherwise(5.7.4)

f(x) = cos ax2 and f(x) = sin ax2 (a > 0) (5.7.5)

f(x) = |x|−12 and f(x) = |x|−

12 e−ax (a > 0). (5.7.6)

Exercise 5.7.4. Find the function f(x) if its Fourier transform is given by

f(p) = e−k|p|, k > 0. (5.7.7)

Exercise 5.7.5. Let u = u(x, y) be a solution to the Laplace equation on the half-plane y ≥ 0satisfying the conditions

∆u(x, y) = 0, y > 0

u(x, 0) = φ(x)

u(x, y)→ 0 as y → +∞ for every x ∈ R (5.7.8)

1) Prove that the Fourier transform of u in the variable x

u(p, y) =1

2π

∫ ∞−∞

u(x, y)e−ipxdx

has the formu(p, y) = φ(p)e−y |p|.

Here φ(p) is the Fourier transform of the boundary function φ(x).

2) Derive the following formula for the solution to the b.v.p. (5.7.8)

u(x, y) =1

π

∫ ∞−∞

y

(x− s)2 + y2φ(s) ds. (5.7.9)

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